# If $B\subset\mathbb{R}$ is a Borel set and $f:B\to\mathbb{R}$ is an increasing function then $f(B)$ is a Borel set

I am trying to prove the following statement:

"If $$B\subset\mathbb{R}$$ is a Borel set and $$f:B\to\mathbb{R}$$ is an increasing function then $$f(B)$$ is a Borel set"

but I have only managed to prove the easier statement

"If $$B\subset\mathbb{R}$$ is a Borel set and $$f:\mathbb{R}\to\mathbb{R}$$ is a strictly increasing function then $$f(B)$$ is a Borel set"

My proof (of the easier statement):

By Inverse function $$f^{-1}:f(\mathbb{R})\to\mathbb{R}$$ of a strictly increasing function $$f:\mathbb{R}\to\mathbb{R}$$ is continuous we have that $$f^{-1}:f(\mathbb{R})\to\mathbb{R}$$ is continuous so $$(f^{-1})^{-1}(\mathbb{R})=f(\mathbb{R})$$ is open hence Borel thus $$f^{-1}$$ is a continuous function defined on a Borel set so it is Borel measurable which implies that $$(f^{-1})^{-1}(B)=f(B)$$ is a Borel set, as desired. $$\square$$

I would like to prove the initial statement but I have been stuck for a while so I would appreciate an hint about how to tackle its proof, thanks.

• Jun 17, 2021 at 11:55
• math.stackexchange.com/questions/3237588/…). This link seems to solve the same problem but in more generality Jun 17, 2021 at 13:11
• monotone functions are borel and the image of a borel set through a borel injection is borel Jun 17, 2021 at 18:57

Here is a self-contained proof. We first note that it suffices to prove:

Claim. Let $$f : \mathbb{R} \to \mathbb{R}$$ be non-decreasing. Then for each Borel set $$B \subseteq \mathbb{R}$$, the set $$f(B)$$ is also Borel.

To prove this, let $$\mathcal{F}$$ be the set of all Borel sets $$B$$ for which $$f(B)$$ is also Borel. Then we check the following properties of $$\mathcal{F}$$:

1. $$\mathcal{F}$$ contains any open intervals.

Indeed, let $$I$$ be an open interval, and consider any connected component $$C$$ of $$\mathbb{R}\setminus f(I)$$.1) We show that $$C$$ cannot be a singleton. Otherwise, write $$C = \{y\}$$. Then we can find sequences $$(a_n)_{n=1}^{\infty}$$ and $$(b_n)_{n=1}^{\infty}$$ in $$I$$ such that

$$\begin{gather*} f(a_1) < f(a_2) < \cdots < y < \cdots < f(b_2) < f(b_1), \\ \lim f(a_n) = y, \qquad \lim f(b_n) = y. \end{gather*}$$

This in particular forces that $$a_1 < a_2 < \cdots < b_2 < b_1$$, and so, both $$a = \lim a_n$$ and $$b = \lim b_n$$ exist in $$I$$ and $$a \leq b$$. Moreover, $$y \leq f(a) \leq f(b) \leq y$$, and so, $$f(a) = f(b) = y$$. This contradicts that $$y$$ lies in the complement of $$f(I)$$, and so, $$C$$ is not singleton as required.

Now this tells that $$\mathbb{R}\setminus f(I)$$ is a disjoint union of nondegenerate intervals, and since there are at most countably many such intervals, their union is a Borel set. Therefore $$f(I)$$ is also a Borel set.

2. If $$A_1, A_2, \dots$$ are in $$\mathcal{F}$$, then both $$\cup_n A_n$$ and $$f(\cup_n A_n) = \cup_n f(A_n)$$ are Borel sets, hence $$\cup_n A_n \in \mathcal{F}$$.

3. If $$A \in \mathcal{F}$$, then $$\mathbb{R}\setminus A \in \mathcal{F}$$.

Indeed, denote by $$D$$ the set of values in $$\mathbb{R}$$ which are attained by $$f$$ at two or more points, i.e.,

$$D = \{y \in \mathbb{R} : \#f^{-1}(\{y\}) \geq 2. \}$$

For each $$y \in D$$ and for each $$a, b \in f^{-1}(\{y\})$$ with $$a < b$$, we find that $$[a, b] \subseteq f^{-1}(\{y\})$$. This tells that each $$f^{-1}(\{y\})$$ is a nondegenerate interval, and so, $$D$$ can contain at most countably many points. Then by noting that $$f(A) \cap f(\mathbb{R}\setminus A) \subseteq D$$, we have

$$f(\mathbb{R})\setminus f(A) \subseteq f(\mathbb{R}\setminus A) \subseteq (f(\mathbb{R})\setminus f(A)) \cup D.$$

In particular, $$f(\mathbb{R}\setminus A) = (f(\mathbb{R})\setminus f(A)) \cup \tilde{D}$$ for some subset $$\tilde{D}$$ of $$D$$. Since all of $$f(\mathbb{R})$$, $$f(A)$$, and $$\tilde{D}$$ are Borel sets, $$f(\mathbb{R}\setminus A)$$ is also a Borel set and hence $$\mathbb{R}\setminus A \in \mathcal{F}$$ as desired.

Conclusion. The above observations tell that $$\mathcal{F}$$ is a $$\sigma$$-algebra containing all the open intervals, and hence, must contain any Borel sets. Therefore the proof is complete. $$\square$$

1) If you are not familiar with the notion of connect components, then here is an alternative way of explaining this for subsets of $$\mathbb{R}$$. Let $$E \subseteq \mathbb{R}$$, and define the relation $$\sim$$ on $$E$$ as follows:

• For each $$x, y \in E$$, we write $$x \sim y$$ if and only if $$\{tx + (1-t)y : t \in [0, 1]\} \subseteq E$$.

In other words, $$x \sim y$$ if and only if either $$x = y$$ or the closed interval between $$x$$ and $$y$$ lie in $$E$$. Then it is not hard to check that

1. $$\sim$$ is an equivalence relation on $$\mathbb{R}$$, and
2. each equivalence class of $$\sim$$ is either a singleton or an interval.
• thank you very much, I have a few questions about your proof; since I am not very familiar with topology could you please tell me why in point (1) you consider a "connected component"? What is it and what is the intuition behind this choice? Also, shouldn't the inequality in (1) be $f(a)\leq y\leq f(b)$? Also, in part (3) you say "For each $y\in D$ and for each $a,b\in f^{-1}(\{y\})$ with $a<b$, we find that $[a,b]\subseteq f^{-1}(\{y\})$": why is that? Thanks Aug 2, 2021 at 14:07
• @lorenzo, For (1), I added a brief explanation to my answer. The idea is that if $f$ is increasing and makes a jump at a point, then that jump will manifest as intervals in the complement $\mathbb{R}\setminus f(I)$. For instance, if we consider $f(x)=\operatorname{sgn}(x)$, then $$\mathbb{R}\setminus f((-1,1))=(-\infty,-1)\cup(-1,0)\cup(0,1)\cup(1,\infty),$$ and the part $(-1,0)\cup(0,1)$ corresponds to the jump of $\operatorname{sgn}(x)$ at $x=0$. Aug 2, 2021 at 17:54
• @lorenzo, for (2), the inequality you mentioned is also true, but it alone does not nail down the values of $f(a)$ and $f(b)$. To explain why we expect $y\leq f(a)\leq f(b)\leq y$ to hold, the idea is that $C$ cannot be a singleton because $C$ corresponds to (at least part of) a jump of $f$, and the zero jump size will imply continuity, i.e., no jump. Now using $(a_n)$ and $(b_n)$ chosen as in the answer, note that $a_n \leq a \leq b \leq b_n$, and so, $f(a_n) \leq f(a) \leq f(b) \leq f(b_n)$. Then letting $n\to\infty$ and $\lim f(a_n)=\lim f(b_n) = y$ will give us the inequality. Aug 2, 2021 at 17:59
• @lorenzo, Finally, the claim mentioned in (3) more or less tells that if you have $f(a) = f(b)$, then the monotonicity of $f$ forces that the graph of $f$ on the interval $[a, b]$ must be flat, i.e., $f(x) = f(a) = f(b)$ for all $x \in [a,b]$. This then forces that $[a,b]$ lies in the inverse image of $\{y\}$ under $f$. Aug 2, 2021 at 18:00
• fantastic answer! Thank you very much! Aug 3, 2021 at 14:37